A positive charge of 1.0 * 10 ^ (-5) C experiences a force of 0.20 N when located at a certain point.What is the electric field intensity at that point ?
E =F/q
Well, if the positive charge is experiencing a force, it must be getting "charged" with some electrifying energy! Haha!
To find the electric field intensity at that point, we can use the formula:
Electric Field Intensity = Force / Charge
So, plugging in the numbers from your question:
Electric Field Intensity = 0.20 N / 1.0 * 10^(-5) C
Calculating it out, we get:
Electric Field Intensity = 20,000 N/C
That's quite an electrifying field! Just make sure you don't get too charged up around it!
To find the electric field intensity at a certain point, you can use the formula:
Electric Field Intensity (E) = Force (F) / Charge (Q)
Given:
Charge (Q) = 1.0 * 10^(-5) C
Force (F) = 0.20 N
Substituting the values into the formula, we get:
E = F / Q
E = 0.20 N / 1.0 * 10^(-5) C
E = 0.20 N / 0.00001 C
E = 20,000 N/C
Therefore, the electric field intensity at that point is 20,000 N/C.
To find the electric field intensity at a certain point, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's Law is given by the equation:
F = k * q1 * q2 / r^2
where:
F is the magnitude of the force between the charges,
k is Coulomb's constant, which is approximately 9.0 * 10^9 Nm^2/C^2,
q1 and q2 are the magnitudes of the two charges, and
r is the distance between the charges.
In this case, we have a positive charge of magnitude 1.0 * 10^(-5) C experiencing a force of 0.20 N. We can rearrange Coulomb's Law to solve for the electric field intensity (E):
E = F / q1
Substituting the given values, we have:
E = 0.20 N / (1.0 * 10^(-5) C)
Calculating this expression gives us the electric field intensity at that point.